Question

The V-I graph for conductor at temperatures $T_1$ and $T_2$ are shown in the figure. ($T_2$ - $T_1$) is proportional to :

• A. $cos2\theta$
• B. $sin2\theta$
• C. $cot2\theta$
• D. $tan2\theta$

Answer 1

The correct option is C $cot2\theta$

We know that-

$R_1=R_o[1+(T_1-T_o\left)\alpha \right]$

$R_2=R_o[1+(T_2-T_o\left)\alpha \right]$

$⟹R_2-R_1=R_o\alpha (T_2-T_1)$

Now, we know that resistance is the slope of $V-I$ curve.

Hence, $R_2=\tan ({90}^o-\theta )=\cot \theta$

and $R_1=\tan \theta$

Hence, $\cot \theta -\tan \theta =R_o(T_2-T_1)\alpha$

$⟹R_o(T_2-T_1)\alpha =\frac{1-{\tan }^2\theta }{\tan \theta }=2\frac{1-{\tan }^2\theta }{2\tan \theta }=\frac{2}{\tan 2\theta }=2\cot 2\theta$

$⟹T_2-T_1\propto \cot 2\theta$

Answer-(C)